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Factorization using Perfect Square Trinomials

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Factorization using Perfect Square Trinomials

What if you had a trinomial expression like $x^2 + 10x + 25$ in which the first and third terms were perfect squares and the second term was twice the product of the square roots of the first and third terms? How could you factor that expression? After completing this Concept, you'll be able to factor perfect square trinomials like this one.

Watch This

For more examples of factoring perfect square trinomials, watch the videos at http://www.onlinemathlearning.com/perfect-square-trinomial.html .

Guidance

We use the square of a binomial formula to factor perfect square trinomials. A perfect square trinomial has the form $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$ .

In these special kinds of trinomials, the first and last terms are perfect squares and the middle term is twice the product of the square roots of the first and last terms. In a case like this, the polynomial factors into perfect squares:

$a^2 + 2ab + b^2 & = (a + b)^2\\a^2 - 2ab + b^2 & = (a - b)^2$

Once again, the key is figuring out what the $a$ and $b$ terms are.

Example A

Factor the following perfect square trinomials:

a) $x^2 + 8x +16$

b) $x^2 - 4x + 4$

c) $x^2 + 14x +49$

Solution

a) The first step is to recognize that this expression is a perfect square trinomial.

First, we can see that the first term and the last term are perfect squares. We can rewrite $x^2 + 8x + 16$ as $x^2 + 8x + 4^2$ .

Next, we check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite $x^2 + 8x + 16$ as $x^2 + 2 \cdot 4 \cdot x + 4^2$ .

This means we can factor $x^2 + 8x + 16$ as $(x + 4)^2$ . We can check to see if this is correct by multiplying $(x + 4)^2 = (x + 4)(x + 4)$ :

$& \quad \quad \quad x + 4\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;x + 4}\\& \quad \quad \ 4x + 16\\& \underline{x^2 + 4x\;\;\;\;\;\;\;\;}\\& x^2 + 8x + 16$

Note: We could factor this trinomial without recognizing it as a perfect square. We know that a trinomial factors as a product of two binomials:

$(x\;\;\;\;)(x\;\;\;\;)$

We need to find two numbers that multiply to 16 and add to 8. We can write 16 as the following products:

$& 16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17\\& 16 = 2 \cdot 8 && \text{and} && 2 + 8 = 10\\& 16 = 4 \cdot 4 && \text{and} && 4 + 4 = 8 \qquad These \ are \ the \ correct \ numbers$

So we can factor $x^2 + 8x + 16$ as $(x + 4)(x + 4)$ , which is the same as $(x + 4)^2$ .

Once again, you can factor perfect square trinomials the normal way, but recognizing them as perfect squares gives you a useful shortcut.

b) Rewrite $x^2 + 4x + 4$ as $x^2 + 2 \cdot (-2) \cdot x + (-2)^2$ .

We notice that this is a perfect square trinomial, so we can factor it as $(x - 2)^2$ .

c) Rewrite $x^2 + 14x + 49$ as $x^2 + 2 \cdot 7 \cdot x + 7^2$ .

We notice that this is a perfect square trinomial, so we can factor it as $(x + 7)^2$ .

Example B

Factor the following perfect square trinomials:

a) $4x^2 + 20x + 25$

b) $9x^2 - 24x + 16$

c) $x^2 + 2xy + y^2$

Solution

a) Rewrite $4x^2 + 20x + 25$ as $(2x)^2 + 2 \cdot 5 \cdot (2x) + 5^2$ .

We notice that this is a perfect square trinomial and we can factor it as $(2x + 5)^2$ .

b) Rewrite $9x^2 - 24x + 16$ as $(3x)^2 + 2 \cdot (-4) \cdot (3x) + (-4)^2$ .

We notice that this is a perfect square trinomial and we can factor it as $(3x - 4)^2$ .

We can check to see if this is correct by multiplying $(3x - 4)^2 = (3x - 4)(3x - 4)$ :

$& \quad \quad \quad 3x - 4\\& \underline{\;\;\;\;\;\;\;\;\;\;\;3x - 4\;\;}\\& \quad \ -12x + 16\\& \underline{9x^2 - 12x\;\;\;\;\;\;\;\;}\\& 9x^2 - 24x + 16$

c) $x^2 + 2xy + y^2$

We notice that this is a perfect square trinomial and we can factor it as $(x + y)^2$ .

Solve Quadratic Polynomial Equations by Factoring

With the methods we’ve learned in the last two sections, we can factor many kinds of quadratic polynomials. This is very helpful when we want to solve them. Remember the process we learned earlier:

1. If necessary, rewrite the equation in standard form so that the right-hand side equals zero.
2. Factor the polynomial completely.
3. Use the zero-product rule to set each factor equal to zero.
4. Solve each equation from step 3.

We can use this process to solve quadratic polynomials using the factoring methods we just learned.

Example C

Solve the following polynomial equations.

a) $x^2 + 12x + 36 = 0$

b) $x^2 - 24x = -144$

Solution

a) Rewrite:

The equation is already in the correct form.

Factor:

Rewrite $x^2 + 12x + 36 = 0$ as $x^2 + 2(6x) + 6^2 = 0$ . We notice that this is a perfect square trinomial and we can factor it as $(x + 6)^2$ .

Set the factor equal to zero:

$x + 6 = 0$

Solve:

$\underline{\underline{x = -6}}$

Check: Substitute each solution back into the original equation.

$& (-6)^2 + 12(-6) + 36 = && \text{Substitute in -6.}\\& 36 + -72 + 36 = && \text{Simplify.}\\& 72 + -72 = 0&& \text{Checks out.}$

b) Rewrite: $x^2 - 24x = -144$ is rewritten as $x^2 - 24x + 144 = 0$

Factor:

$x^2 - 24x + 144 =x^2+2(-12)x+(-12)^2=(x-12)^2$

Set the factor equal to zero:

$x - 12 = 0$

Solve:

$\underline{\underline{x = 12}}$

Check: Substitute the solution back into the original equation.

$& (12)^2 - 24(12) + 144 = && \text{Substitute in 12.}\\& 144 - 288+144 = && \text{Simplify.}\\& 288- 288 = 0 && \text{Checks out.}\\$

Watch this video for help with the Examples above.

Vocabulary

• A perfect square trinomial has the form

$a^2+2ab+b^2=(a+b)^2 \qquad \text{or} \qquad a^2-2ab+b^2=(a-b)^2.$

Guided Practice

Solve the following polynomial equations:

a) $x^2 + x + 0.25 = 0$

b) $x^2 - 81 = 0$

Solution

a) $x^2 + x + 0.25 = 0$

Rewrite: The equation is in the correct form already.

Factor: Rewrite $x^2 + x + 0.25 =0$ as $x^2 + 2 \cdot (0.5)x + ( 0.5)^2$ .

We recognize this as a perfect square. This factors as $(x +0.5)^2 = 0$ or $(x +0.5)(x +0.5) = 0$

Set the factor equal to zero:

$x +0.25 = 0$

Solve:

$\underline{\underline{x = -0.5}}$

Check: Substitute the solution back into the original equation.

$(-0.5)^2 + -0.5 + 0.25&= \quad \quad \text{Substitute in -0.5}\\0.25+ -0.5 + 0.25&= \quad \quad \text{Simplify.}\\0.5-0.5&=0 \quad \quad \text{Checks out.}$

b) $x^2 - 81 = 0$

Rewrite: this is not necessary since the equation is in the correct form already

Factor: Rewrite $x^2 - 81$ as $x^2 - 9^2$ .

We recognize this as a difference of squares. This factors as $(x - 9)(x + 9) = 0$ .

Set each factor equal to zero:

$x - 9 = 0 && \text{or} && x + 9 = 0$

Solve:

$\underline{\underline{x = 9}} && \text{or} && \underline{\underline{x = - 9}}$

Check: Substitute each solution back into the original equation.

$& x = 9 && 9^2 - 81 = 81-81 = 0 && \text{checks out}\\& x = -9 && (-9)^2 - 81 = 81 - 81 = 0 && \text{checks out}$

c) $x^2 + 20x + 100 =0$

Rewrite: this is not necessary since the equation is in the correct form already

Factor: Rewrite $x^2 + 20x + 100$ as $x^2 + 2 \cdot 10 \cdot x + 10^2$ .

We recognize this as a perfect square. This factors as $(x + 10)^2 =0$ or $(x + 10)(x + 10)=0$

Set each factor equal to zero:

$x + 10 =0 && \text{or} && x + 10 = 0$

Solve:

$\underline{\underline{x = -10}} && \text{or} && \underline{\underline{x = -10}} \quad \quad \text{This is a double root.}$

Check: Substitute each solution back into the original equation.

$x = 10 && (-10)^2 + 20(-10) + 100 = 100 -200 + 100 =0 && \text{checks out}$

Practice

Factor the following perfect square trinomials.

1. $x^2 + 8x + 16$
2. $x^2 - 18x + 81$
3. $-x^2 + 24x -144$
4. $x^2 + 14x + 49$
5. $4x^2 - 4x + 1$
6. $25x^2 + 60x + 36$
7. $4x^2 - 12xy + 9y^2$
8. $x^4 + 22x^2 + 121$

Solve the following quadratic equations using factoring.

1. $x^2 - 11x + 30 =0$
2. $x^2 + 4x =21$
3. $x^2 + 49 =14x$
4. $x^2 - 64 = 0$
5. $x^2 - 24x + 144 = 0$
6. $4x^2 - 25 = 0$
7. $x^2 + 26x = -169$
8. $-x^2 - 16x - 60 = 0$